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I've recently come across an integral of the form $$\int_{0}^{\infty} \sinh(\eta)^2\int_{-1}^{1}f_1(a\cosh(\eta)+b\sinh(\eta)x)f_2(a\cosh(\eta)-b\sinh(\eta)x)dxd\eta$$ Where $a>b$, and $f_1$ and $f_2$ are some real valued functions that decay very rapidly (safe to assume $f_i(x) \sim O(e^{-\alpha_i x})$ for $\alpha_i>0$). I am trying to compute this in terms of their Fourier transforms $\hat{f_i}$, for which I use: $$f_i(x)=\int_{-\infty}^{\infty}\hat{f_i}(k)e^{ikx}dk$$

I also am only interested in $f_i(x)$ for $x\geq 0$, so we can assume for the sake of computing the Fourier transform $f_i(-x)=f_i(x)$

Plugging this in to the above, we get:

$$\int_{0}^{\infty} \sinh(\eta)^2\int_{-1}^{1}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{f_1}(k_1)\hat{f_2}(k_2)e^{ia\cosh(\eta)(k_1+k_2)}e^{ib\sinh(\eta)(k_1-k_2)x}dk_1dk_2dxd\eta$$

I want to reduce this to a double integral, and have come very close but am encountering some issues. I will show why I believe we can reduce this to a double integral--the following is not rigorous in any way. Some of my steps are unjustified, with the hope that the resulting integral will converge to the same value.

Changing the order of integration (unjustified):

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{f_1}(k_1)\hat{f_2}(k_2)\int_{0}^{\infty} \sinh(\eta)^2e^{ia\cosh(\eta)(k_1+k_2)}\int_{-1}^{1}e^{ib\sinh(\eta)(k_1-k_2)x}dxd\eta dk_1dk_2$$

The inntermost integral is easy to compute: $$\int_{-1}^{1}e^{ib\sinh(\eta)(k_1-k_2)x}dx=\frac{e^{ib\sinh(\eta)(k_1-k_2)}-e^{-ib\sinh(\eta)(k_1-k_2)}}{ib\sinh(\eta)(k_1-k_2)}$$

This allows us to complete the integral over $\eta$ as well: $$\int_{0}^{\infty} \sinh(\eta)^2e^{ia\cosh(\eta)(k_1+k_2)}\int_{-1}^{1}e^{ib\sinh(\eta)(k_1-k_2)x}dxd\eta =\\ \frac{1}{ib(k_1-k_2)}\int_{0}^{\infty} \sinh(\eta)e^{ia\cosh(\eta)(k_1+k_2)}[e^{ib\sinh(\eta)(k_1-k_2)}-e^{-ib\sinh(\eta)(k_1-k_2)}]d\eta= \\ \frac{1}{ib(k_1-k_2)}\int_{-\infty}^{\infty} \sinh(\eta)e^{ia\cosh(\eta)(k_1+k_2)}e^{ib\sinh(\eta)(k_1-k_2)}d\eta =\\ \frac{-1}{b(k_1-k_2)^2} \frac{\partial}{\partial b}\int_{-\infty}^{\infty} e^{ia\cosh(\eta)(k_1+k_2)}e^{ib\sinh(\eta)(k_1-k_2)}d\eta$$ To my luck, I was able to find a remarkable identity to compute this: $$H_0^{1}(\sqrt{a^2-b^2})=\frac{1}{i\pi}\int_{-\infty}^{\infty} e^{ia\cosh(\eta)}e^{ib\sinh(\eta)}d\eta$$ $$\frac{\partial}{\partial b}H_0^{1}(\sqrt{a^2-b^2})=\frac{b}{\sqrt{a^2-b^2}}H_1^{1}(\sqrt{a^2-b^2})$$ Where $H_{\nu}^{1}$ is the Hankel function of the first type.

Unfortunately, the identity only holds for $\Im(a \pm b)>0$. However, $H_{\nu}^{1}(\sqrt{a^2-b^2})=J_{\nu}(\sqrt{a^2-b^2})+iY_{\nu}(\sqrt{a^2-b^2})$ is still defined for $a>b,a,b\in \mathbb{R}$. So, let me just pretend that we are justified to use this here. Using this formula, we observe:

$$\frac{-1}{b(k_1-k_2)^2} \frac{\partial}{\partial b}\int_{-\infty}^{\infty} e^{ia\cosh(\eta)(k_1+k_2)}e^{ib\sinh(\eta)(k_1-k_2)}d\eta= \\ \frac{-i\pi}{b(k_1-k_2)^2} \frac{\partial}{\partial b}H_0^{1}(\sqrt{a^2(k_1+k_2)^2-b^2(k_1-k_2)^2})= \\ \frac{-i\pi}{\sqrt{a^2(k_1+k_2)^2-b^2(k_1-k_2)^2}} H_1^{1}(\sqrt{a^2(k_1+k_2)^2-b^2(k_1-k_2)^2})$$

Finally, plugging back into the initial expression, we obtain:

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{f_1}(k_1)\hat{f_2}(k_2) \frac{-i\pi}{\sqrt{a^2(k_1+k_2)^2-b^2(k_1-k_2)^2}} H_1^{1}(\sqrt{a^2(k_1+k_2)^2-b^2(k_1-k_2)^2}) dk_1dk_2$$

The issue with this is the argument of $H_1^{1}$ can be negative, landing us on a branch cut, and numerical tests show this result does not hold. At this point, I am unsure how to deal with this.

I would greatly appreciate if anyone with more knowledge than me could guide me on how to procede/modify my efforts, or tell me if this route likely will not work.

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  • $\begingroup$ $$ \int_0^\infty \sinh^2\eta\!\int_{-1}^1 f_1f_2\,dx\,d\eta = \int_{-\infty}^\infty\!\int_{-\infty}^\infty \hat f_1(k_1)\,\hat f_2(k_2)\, \frac{2}{b(k_1-k_2)} \int_1^\infty e^{i a(k_1+k_2)u}\, \sin\!\bigl(b(k_1-k_2)\sqrt{u^2-1}\bigr) \,du\;dk_1\,dk_2. $$ $\endgroup$ Commented Jun 21 at 5:05
  • $\begingroup$ @Frank my goal is to eliminate the innermost integral, if possible $\endgroup$ Commented Jun 21 at 5:48
  • $\begingroup$ You may be interested in the proof of the following identity (Gradshteyn \& Rhyzhik (4.620.5)) \begin{align} & \int_0^{\infty} d x \int_0^\pi f(p \cosh x+q \cos \omega \sinh x) \sinh ^2 x \sin \omega d \omega=\\ &\hspace{2cm} =2\int_0^{\infty} f\left(\operatorname{sign} p \sqrt{p^2-q^2} \cosh x\right) \sinh ^2 x d x \\ & \hspace{6cm}\left[\lim _{x \rightarrow+\infty} f(x)=0\right] \end{align} $\endgroup$ Commented Jun 24 at 21:12
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    $\begingroup$ @PaulEnta amazing find! This is so close to what I need. Hopefully their derivation can be modified for my case. Thank you for your help! $\endgroup$ Commented Jun 25 at 6:56
  • $\begingroup$ Yes, I hope that symmetry will help to extend the derivation to your case. $\endgroup$ Commented Jun 25 at 7:20

1 Answer 1

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To avoid determination issues in the evaluation of the intermediate integral \begin{align} I(k_1,k_2)&=\int_{-\infty}^{\infty} e^{ia\cosh(\eta)(k_1+k_2)}e^{ib\sinh(\eta)(k_1-k_2)}d\eta\\ &=\int_{-\infty}^\infty\exp\left( i(a_+\cosh\eta+b_-\sinh\eta) \right)\,d\eta \end{align} (with the notations $a_+=a(k_1+k_2)$ and $b_-=a(k_1-k_2)$) we treat differently the situations depending on the relative values of $a_+$ and $b_m$. .

When $a_+>b_-$, one can define a real parameter $s=\tanh^{-1}\frac{b_-}{a_+}$ such that \begin{align} a_+\cosh\eta+b_-\sinh\eta&=\sqrt{a_+^2-b_-^2}\left( \frac{a_+\cosh\eta}{\sqrt{a_+^2-b_-^2}}+\frac{b_-\sinh\eta}{\sqrt{a_+^2-b_-^2}} \right)\\ &=\sqrt{a_+^2-b_-^2}\left( \cosh\eta\cosh s+\sinh\eta\sinh s \right)\\ &=\sqrt{a_+^2-b_-^2}\cosh(\eta-s) \end{align} By plugging this representation into the integral above, it comes \begin{align} I(k_1,k_2)&=\int_{-\infty}^\infty\exp\left( i\sqrt{a_+^2-b_-^2}\cosh(\eta-s) \right)\,d\eta\\ &=\int_{-\infty}^\infty\exp\left( i\sqrt{a_+^2-b_-^2}\cosh(t) \right)\,dt\\ &=2\int_0^\infty\cos\left( \sqrt{a_+^2-b_-^2}\cosh(t) \right)\,dt+2i\int_0^\infty\sin\left( \sqrt{a_+^2-b_-^2}\cosh(t) \right)\,dt \end{align} where the integration variable $\eta$ was shifted $t=\eta-s$ and parity of the integrand was used. Now, using DLMF 10.9.9: \begin{align} J_{0}\left(x\right)&=\frac{2}{\pi}\int_{0}^{\infty}\sin\left(x\cosh t\right)\,% \mathrm{d}t\\ Y_{0}\left(x\right)&=-\frac{2}{\pi}\int_{0}^{\infty}\cos\left(x\cosh t\right)\,% \mathrm{d}t \end{align} when $x>0$, it can be deduced \begin{align} I(k_1,k_2)&=-\pi Y_0\left( \sqrt{a_+^2-b_-^2} \right)+\pi iJ_0\left( \sqrt{a_+^2-b_-^2} \right)\\ &=i\pi H_0^{(1)}\left( \sqrt{a_+^2-b_-^2} \right) \end{align}

When $b_->a_+$, using the same method, it can be shown that \begin{align} I(k_1,k_2)&=\int_{-\infty}^\infty\exp\left( i\sqrt{b_-^2-a_+^2}\sinh(t) \right)\,dt\\ &=2\int_0^\infty\cos\left( \sqrt{b_-^2-a_+^2}\sinh(t) \right)\,dt \end{align} (as the odd term gives a vanishing contribution). With DLMF 10.32.7: \begin{equation} K_{\nu}\left(x\right)=\sec\left(\tfrac{1}{2}\nu\pi\right)\int_{0}^{\infty}\cos% \left(x\sinh t\right)\cosh\left(\nu t\right)\,\mathrm{d}t \end{equation} (for $\left|\Re \nu\right|<1,x>0$), it comes \begin{equation} I(k_1,k_2)=2K_0\left( \sqrt{b_-^2-a_+^2} \right) \end{equation} These results seem numerically correct.

However, differentiation wrt $b$ which is required in OP's approach leads to divergent integrals near the point $a_+=b_-$. This is probably related to the restriction $|\Re\nu|<1$ in 10.32.7 and 10.9.8. Indeed the expressions in terms of $K_1$ and $H_1^{(1)}$ could have been directly obtained from these identities (violating the condition on $\nu$) instead of using the differentiation wrt $b$. This problem is likely due to the change of the order of integration.

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    $\begingroup$ Thank you for this great answer! It looks sensible but I am unable to verify it--it looks like the singularity makes integrating very slow. I am wondering if you managed to verify the final expression you gave $\endgroup$ Commented Jun 23 at 18:44
  • $\begingroup$ No, I didn't check the final integral. But you are right, these integrals do not converge near the point $a_+=b_-$ (neither the one you obtained which is completely similar). This is probably related to the restriction $|\Re\nu|<1$ in 10.32.7 and 10.9.8. Indeed the expressions in terms of $K_1$ and $H_1^{(1)}$ could have been directly obtained instead using the differentiation wrt $b$. Now, I don't see how to solve this problem which is likely due to the change of the order of integration. $\endgroup$ Commented Jun 23 at 21:38

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